Question

# $△\mathrm{ABC}$ is an isosceles triangle right angled at $\mathrm{C}$. Prove that ${\mathrm{AB}}^{2}=2{\mathrm{AC}}^{2}.$

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Solution

## Proving that ${\mathrm{AB}}^{2}=2{\mathrm{AC}}^{2}$:$\mathrm{\Delta ABC}$is an isosceles triangle right angled at $C$. In $\mathrm{\Delta ACB},\angle \mathrm{C}=90°$ $\mathrm{AC}=\mathrm{BC}$ [By isosceles triangle property]${\mathrm{AB}}^{2}={\mathrm{AC}}^{2}+{\mathrm{BC}}^{2}$ [By Pythagoras theorem] ${\mathrm{AB}}^{2}={\mathrm{AC}}^{2}+{\mathrm{AC}}^{2}$ [Since, $\mathrm{AC}=\mathrm{BC}$]$⇒$${\mathrm{AB}}^{2}=2{\mathrm{AC}}^{2}$Hence we proved that ${\mathrm{AB}}^{2}=2{\mathrm{AC}}^{2}$

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